Question #33891

1 Answer
Jan 5, 2017

sf(102.5color(white)(x)s)

Explanation:

I will consider the motion in two stages.

1. The Powered Stage

To get the final velocity at the end of the powered stage we can use:

sf(v=u+at)

This becomes:

sf(v=0+((9.81)/(3))xx60=196.2color(white)(x)"m/s")

To find the height h reached we can use:

sf(v^2=u^2+2as)

This becomes:

sf(196.2^2=0+(2xx9.81)/(3)xxh)

:.sf(h=(3.849xx10^4xx3)/(2xx9.81)=588.6color(white)(x)m)

2. The Unpowered Stage

The rocket is now moving under the influence of gravity.

We can use:

sf(s=ut+1/2at^2)

I will use the convention that "up is positive".

The equation becomes:

sf(-588.6=196.2t-1/2xx9.81xxt^2)

:.sf(4.9t^2-196.2t-588.6=0)

Applying the quadratic formula:

sf(t=(196.2+-sqrt(37,094.76-[4xx4.9xx(-588.6)]))/(9.8)

sf(t=(196.2+-220.5)/(9.8))

Ignoring the -ve root:

sf(t=42.5color(white)(x)s)

To get the total time of flight (sf(t_("tot"))), we add this to the time taken for the powered part of the flight which we know is 1 minute.

:.sf(t_("tot")=60+42.5=102.5color(white)(x)s)